A pilot is flying from City A to City B which is 300 km [NW]. If the plane will encounter a constant wind of 80 km/h from the north and the schedule insists that he complete his trip in 0.75 h, what air speed and heading should the plane have?

Answer 1

V=d/t = 300/0.75=400 km/h[135o]. Without
wind.

V = 400[135] + 80[270o]

X == 400*cos135 = -282.8 km/h.
Y = 400*sin135 + 80*sin270 = 202.8 km/h.

Tan Ar = Y/X = 202.8/-282.8 = -0.71727
Ar = -35.65o = Reference angle.
A = -35.65 + 180 = 144.3o

V = X/cosA = -282.8/cos144.3 = 348 km/h.

(144.3-135) = 9.3o Off due to wind.

Heading = 135 - 9.3 = 125.7o
V=400 + (400-348) = 400 + 52 = 452 km/h